Manipal University 2008-1st Sem B.Tech Chemical Science and Technology MIT - Manipal -B.E - ENGGCHEMISTRY.(CHM 101/102) - Question Paper
MANIPAL INSTITUTE OF TECHNOLOGY MANIPAL UNIVERSITY
I SEMESTER BE DEGREE EXAMINATION NOVEMBER 2008
a
INSPIRED BY LIFE
SUBJECT: ENGG. CHEMISTRY.(CHM 101/102)
Scheme of Valuation _[Max. Marks: 50
Time: 3 Hrs]
Note: Answer any FIVE full questions
1. a) Give reason
i) E = Eo - 0.0592 log [Cl-]----------/M Explanation----------- /M
ii) Two reasons------------------------1 M
b) i) Defination------1M Derivation of Mn & Mw --- -2M
ii) One major difference with an example---------------/M + -/M
c) i) Diagram-----/M Conditions like Temp, Pressure & Catalyst1M
Process-----1M Merits ( any two)--------------/M
ii) Calculation of G.C.V L = ( W + w ) ( t2-ti)/x
= (1458 + 456) ( 19.2-13,25)/0.945
= 12.051 k cal g-1 /M
LCV = 12051 - 8x0.09x586
= 11.629 k cal g-1 /M
2. a) Give reason
i) The secondary reaction taking place in a dry cell is not a electrochemical reaction It is the conversion of Zn into diammine Zn (II) chloride. So it do not contribute towards the overall
NH4+(aq) + OH-(aq) NH3(g) + H2O (l)
Zn2+(aq)+2NH3(s)+ 2 Cl - [Zn(NH3)2 Ch] 1M
ii) Due to overcharging electrolysis of water will take place which results in the accumulation of hydrogen and oxygen leading to explosion of the cell. If kept in a partially discharged condition sulphatation will take place. 1M
b) i)Explanation of function of anodic inhibitor with with an example 1M
Explanation of functions of 2 types of cathodic inhibitor with an example each
2M
c) i) Definition of electroless plating /M
The catalytic metals such as Ni, Co, Steel, Pd etc., do not require any surface preparation before electro less plating on them. Non catalytic metals such as Cu, brass, Ag etc., need activation. This can be done by dipping in palladium chloride solution.Non conductors like glass, plastics, etc., are first activated in a solution of SnCl2 and HCl.After rinsing its immersed in a solution of PdCl2 and HCl. This treatment yields a thin layer of Pd on the treated surface.
1/ M
|
O O HOC(CH2)4C OH + H2N~(CH2)fiNH2 adipic acid hexameihylene diamine o O O |
o o II II -Oc (CH2)4cO- H,N (CH,), NH, nylon salt |
c (CH2)4C-|-NH (CH2)(1NHc (CH2)4c NH CCH2)6NH
I n
poly{hexamcihylene adipamide), called nylon 6,6
(i) It is used as a plastic as well as fibre.
(ii) This is used to produce tyre cord.
(iii)It is used to make mono filaments and roaps.
(iv)Nylon 6,6 is used to manufacture articles like brushes and bristles.
Preparation of SBR
H2C=CH
O
H2CCH=CH CH2-k H2CCH
J x
o
n x
H2C=CHCH =CH2
+ n
n
Uses:
Mainly used for the manufacture of motor tyres.
Other uses of this elastomer are floor tiles, shoe soles, gaskets, foot-wear components, wire and cable insulations, carpet backing, adhesives, tank-lining, etc.,
3. a) Give reason:
i) Mentioning 2 types of cathodic reactions in aerated acid soulutions M Equations M
ii) Reason with proper explanation regarding passivating potential 1M
b) i) Development of electrode potential in a glass electrode:
The glass is a partially hydrated aluminosilicate containing sodium or calcium ions. The hydration of a pH sensitive glass membrane involves an ion-exchange reaction between singly charged cations in the interstices of the glass lattice and protons from the solution. The ion-exchange reaction can be written as
H+ + Na+ <=====> Na+ + H+
Solution glass solution glass M
Explanation for development of potential M.
Development of electrode potential of glass electrode:
The overall potential of the glass electrode has three components.
The boundary potential b, which varies with the pH of the analyte solution. It is made up of two potentials, E1 & E2 which develop at the two surface of the glass membrane i.e. the potential developed at the inner glass surface & the potential developed at the outer glass surface.
Eb = E1-E2 _(1)
Where Eb is the boundary potential
E1 = potential developed at the interface between the exterior of the glass and the analyte solution
E2 = Potential developed at the interface between the internal solution and the interior of the glass. The boundary potential is related to the concentration of hydrogen ion in each of the solution by the Nernst-like equation.
Eb = E1 - E2 = 0.0592 log Cl / C2 _(2)
Where C1 = concentration of the analyte solution C2 = concentration of the internal solution For a glass pH electrode the hydrogen ion concentration of the internal solution is held constant. So eqn. (2) simplifies to
Eb = K + 0.0592 log Ci_
K - 0.0592 pH where K = - 0.0592 log C 2
The boundary potential is then a measure of the hydrogen ion concentration of the
external solution.
2) The potential of the internal Ag/AgCl reference electrode. E Ag/AgCl.
3) A small unpredictable contribution called the asymmetry potential, E asym. The sources of the symmetry potential include the following.
(i) Differing conditions of strain in the two glass surfaces during manufacture
(ii) Mechanical abrasion on the on the outer surface during use
(iii) Chemical etching of the outer surface during use. /M
The asymmetry potential changes slowly with time. The glass electrode potential can be written in the equation form as
EG Eb + EAg/ AgCl + E asym _ (4)
Substitution of eqn - (3) for Eb, gives Eg = K+0.0592 log C1 + EAg/AgCl + E asym
= K - 0.0592 log pH + E Ag/AgCl + E asym_(5)
or Eg = Eog - 0.0592PpH _(6)
where E G = K + EAg/AgCl + E asym.
a combination of three constant terms = constant /M
ii) pH = Ecell / 0.0592
= 0.34/0.0592 = 5.74.
1M
c) i) Mechanism of petroleum knocking 1M
Prevention 1M
ii) Explanation for suspension polymerization 1/M Any two differences between suspension and emulsion polymerization /M
4. a) Reason:
i) Lower in C.V due to volatile matter /M Lower in C.V due to ash /M
ii) Reason 1M
b) i) Cell Scheme:
Cd/Cd(OH)2,KOH,Ni(OH)2, Ni(OH)3/Ni /M
Electrode reactions Anode:
Cd(OH)2(s)+2e- Cd(s) +2OH'(aq)
2Ni(OH)2(S) +2OH (aq)2Ni(OH)3(s)+2e 2Ni(OH)2(s)+Cd(OH)2(s)2Ni(OH)3(s)+ Cd(s)
AM
AM
AM
1M
1M
2M AM 1 AM
Net:
ii) Any four requirements of the fuel cell Any two merits and demerits
c) i)Explanation of caustic embrittlement with equations
ii) Definition of Tg
Comparison of Tg values with structure
5. a)Give reason:
i) Reason with proper explanation
1M
1M
ii) Reason with suitable explanation
b) i) AG = -nFE However all electrochemical reactions are accompanied by
decrease in free energy of the system. Thus , -AG = -nFE Hence e.m.f of the cell is positive 1M
AH = nF[T(5 E/ 5T)p -E]
= 2 X 96500 ( -2.952x10-3 x298 - 0.7653)
= -2 x 96500x 1.6444
= -317.48 k J K-1 . AM
AS = nF (5E / 5T) p
= -2 x 96500x 2. 952x10-3
= -0.5695 kJ AM
ii) Stucture property relationship with respect ot plastic deformation and Chemical resistance 2M
c) i) Detailed procedure for ultimate analysis of coal of determination of carbon and hydrogen with calculation steps. 2M
Percentage of nitrogen = Volume of acid used * Normality of acid *1.4
Weight of coal taken
= 43.75 x 0.1x1.4/ 1.56
= 3.92% 1M
Percentage of sulphur:
Wt. of BaSOobtained * 32 * 100 Wt. of coal taken * 233
= 0.1755 x32x100/ 2.60 x 233
= 0.9270 % 1M
6. a) Give reason
i) Explanation with equation 1M
ii) Explanation for differential areation corrosion 1M
b) i) Explanation for the effect of two parameters on the rate of corrosion 1 mark each 2M
ii) Tacticity : Def M
Three different stuctural configuration of polypropylene 1 M
c) Origin of single electrode potential. 1/2M Explanation of the galvanization 2M Reason M
|
Attachment: |
| Earning: Approval pending. |
